RD Sharma Class 12 Exercise 3.14 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.14 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:49 AM IST

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.14

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Inverse Trigonometric Functions Exercise 3.14 Question 1(i)

Answer: $\frac{-7}{17}$
Hints: First we will solve for $2 \tan ^{-1} \frac{1}{5}$ and we use the formula
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right),-1<x<1$
Given: $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$
Explanation:
$\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ ..............(1)
Let us first solve for $2 \tan ^{-1} \frac{1}{5}$
$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}$ $\left[\because-1<\frac{1}{5}<1\right]$
$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right\}$
$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{\left(\frac{24}{25}\right)}\right\}$
$=\tan ^{-1}\left\{\frac{2}{5} \times \frac{25}{24}\right\}$
$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{5}{12}\right)$................(2)
Now, we know that $\tan \frac{\pi}{4}=1$
$\frac{\pi}{4}=\tan ^{-1}(1)$
Now from equation (1), (2) and (3)
We get $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12} \times 1}\right)\right\}$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \\ \text { forxy }>-1 \end{array}\right]$
$=\tan \left\{\tan ^{-1}\left(\frac{\frac{-7}{12}}{\frac{17}{12}}\right)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{-7}{12} \times \frac{12}{17}\right)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{-7}{17}\right)\right\}$ $\left[\because \tan \left(\tan ^{-1} \theta\right)=\theta\right]$$=\frac{-7}{17}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question

Answer: $\frac{4-\sqrt{7}}{3}$
Hints: First we will convert $\sin ^{-1}\left(\frac{3}{4}\right)$ into $\tan ^{-1}$
Given: $\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right)$
Explanation:
Let$\sin ^{-1}\left(\frac{3}{4}\right)=\theta$ ......(1)
$\sin \theta=\frac{3}{4}=\frac{A B}{A C}$
In$\triangle A B C$
$A B^{2}+B C^{2}=A C^{2}$(By using Pythagoras theorem)
$\begin{aligned} &3^{2}+B C^{2}=4^{2} \\ &9+B C^{2}=16 \\ &B C^{2}=16-9 \\ &B C^{2}=7 \\ &B C=\sqrt{7} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{B C}{A C} \\ &\cos \theta=\frac{\sqrt{7}}{4} \end{aligned}$
Now
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\frac{\sqrt{7}}{4}}{1+\frac{\sqrt{7}}{4}}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{4-\sqrt{7}}{4}}{\frac{4+\sqrt{7}}{4}}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$
$\frac{\theta}{2}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$
$\frac{1}{2} \sin ^{-1} \frac{3}{4}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$ from equation (1)
Now, $\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)$
$=\tan \left(\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)\right)$ from equation (2)
$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$
On rationalizing we get,
$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}} \times \frac{4-\sqrt{7}}{4-\sqrt{7}}}$
$=\sqrt{\frac{(4-\sqrt{7})^{2}}{16-7}}$
$\begin{aligned} &=\sqrt{\frac{(4-\sqrt{7})^{2}}{9}} \\ &=\frac{(4-\sqrt{7})}{3} \end{aligned}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 3

Answer: $\frac{1}{\sqrt{10}}$
Hint: First we will convert $\cos ^{-1} \frac{4}{5}$ into $\sin ^{-1}$
Given: $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$
Explanation:
Let$\cos ^{-1} \frac{4}{5}=\theta$
$\cos \theta=\frac{4}{5}$
we know that
$\begin{aligned} &\cos \theta=1-2 \sin ^{2} \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \end{aligned}$
$\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\frac{4}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{\frac{1}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1}{10}} \\ &\sin \frac{\theta}{2}=\frac{1}{\sqrt{10}} \\ &\frac{\theta}{2}=\sin ^{-1} \frac{1}{\sqrt{10}} \end{aligned}$ from equation (1)
$\frac{1}{2} \cos ^{-1} \frac{4}{5}=\sin ^{-1} \frac{1}{\sqrt{10}}$ ......(2)
Now
$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ from equation (2)
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$ $\left[\because \sin \left[\sin ^{-1}(\theta)\right]=\theta\right]$
$=\frac{1}{\sqrt{10}}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 4

Answer: $\frac{37}{26}$
Hints: we will convert $2 \tan ^{-1} \frac{2}{3}$ into $\sin ^{-1}$ and $\tan ^{-1} \sqrt{3}$ into $\cos ^{-1}$
Given: $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$
Solution:
$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ ..................(1)
First we will solve for $2 \tan ^{-1} \frac{2}{3}$
We know that $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad-1<x<1$
$\tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{2 \times \frac{2}{3}}{1+\left(\frac{2}{3}\right)^{2}}\right)$ $-1<\frac{2}{3}<1$
$\begin{aligned} &=\sin ^{-1}\left(\frac{\frac{4}{3}}{1+\frac{4}{9}}\right) \\ &=\sin ^{-1}\left(\frac{\frac{4}{3}}{\frac{13}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{3} \times \frac{9}{13}\right) \end{aligned}$
$2 \tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{12}{13}\right)$ .................(2)
Now Let $\tan ^{-1} \sqrt{3}=\theta$ .................(3)
$\tan \theta=\frac{\sqrt{3}}{1}=\frac{P}{B}$
By Pythagoras theorem:
$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &=\sqrt{3}^{2}+1^{2} \\ &=3+1 \\ &=4 \\ &H=\sqrt{4} \\ &H=2 \\ &\cos \theta=\frac{B}{H} \\ &\cos \theta=\frac{1}{2} \end{aligned}$
$\theta=\cos ^{-1}\left(\frac{1}{2}\right)$ from equation.........(2)
$\tan ^{-1} \sqrt{3}=\cos ^{-1}\left(\frac{1}{2}\right)$ ..................(4)
Now from equation (1)
$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$
Putting the value of $2 \tan ^{-1} \frac{2}{3}$ and $\tan ^{-1} \sqrt{3}$ from equations (2) and (4) respectively
$\begin{aligned} &=\sin \left(\sin ^{-1} \frac{12}{13}\right)+\cos \left(\cos ^{-1} \frac{1}{2}\right) \\ &=\frac{12}{13}+\frac{1}{2} \\ &=\frac{24+13}{26} \\ &=\frac{37}{26} \end{aligned}$

Inverse Trigonometric Functions Exercise 3.14 Question 2

Answer: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Hints: First we will convert $\sin ^{-1} \frac{3}{5}$ into $\tan ^{-1}$ and after that we use the formula of $2 \tan ^{-1} x$
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \quad-1<x<1$
Given: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Explanation:
$2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
First we will solve for $\sin ^{-1} \frac{3}{5}$
let $\sin ^{-1} \frac{3}{5}=\theta$ ........................(1)
$\sin \theta=\frac{3}{5}=\frac{P}{H}$
By Pythagoras theorem
$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$
$\tan \theta=\frac{P}{B}$
$\tan \theta=\frac{3}{4} \quad \theta=\tan ^{-1} \frac{3}{4}$
From Equation (1)
$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ ..............(2)
Now L.H.S:
$2 \sin ^{-1} \frac{3}{5}$
$=2 \tan ^{-1} \frac{3}{4}$ from equation (2)
$=\tan ^{-1}\left[\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right]$ $\left[-1<\frac{3}{4}<1\right.$
$=\tan ^{-1}\left[\frac{\frac{3}{2}}{1-\frac{9}{16}}\right]$
$=\tan ^{-1}\left[\frac{\frac{3}{2}}{\frac{7}{16}}\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{3}{2} \times \frac{16}{7}\right] \\ &=\tan ^{-1}\left[\frac{24}{7}\right] \\ &=R \cdot H . S \end{aligned}$
Hence $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 2

Answer: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$
Hints: First we will solve for $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$ then we will convert it into $\cos ^{-1} \text { and } \sin ^{-1}$ respectively.
Given:
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$
First we will solve for $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{1-\frac{1}{18}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \\ &=\tan ^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) \end{aligned}$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1} \frac{1}{2}$
$=\frac{1}{2} \times 2 \tan ^{-1} \frac{1}{2}$ (On multiplying and dividing by 2)
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \times \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right) \quad\left(\begin{array}{l} \because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ 0 \leq x<\infty \end{array}\right)$
$\begin{aligned} &=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right) \\ &=\frac{1}{2} \cos ^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right) \end{aligned}$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$ .......................(1)
Now,let $\cos ^{-1} \frac{3}{5}=\theta$ ....................(2)
$\begin{aligned} &\cos \theta=\frac{3}{5} \\ &\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta} \end{aligned}$ $\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$
$\begin{aligned} \sin \theta &=\sqrt{1-\cos ^{2} \theta} \\ \sin \theta &=\sqrt{1-\left(\frac{3}{5}\right)^{2}} \\ &=\sqrt{1-\frac{9}{25}} \\ &=\sqrt{\frac{25-9}{25}} \\ &=\sqrt{\frac{16}{25}} \end{aligned}$
$\begin{aligned} &\sin \theta=\frac{4}{5} \\ &\theta=\sin ^{-1} \frac{4}{5} \end{aligned}$
from equation (2)
$\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}$ .........(3)
From equation (1) and (2) we get
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 3

Answer: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$
Hint: First we will multiply in numerator and denominator by 2 in L.H.S so that we can use the formula of $2 \tan ^{-1} x$
Given: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$
Explanation:
L.H.S:$\tan ^{-1} \frac{2}{3}$
On multiply in numerator and denominator by 2.
$=\frac{1}{2} \times 2 \tan ^{-1} \frac{2}{3}$
$=\frac{1}{2} \times \tan ^{-1}\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{2}{3}<1 \end{array}\right]$
$\begin{aligned} &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{1-\frac{4}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{\frac{5}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right) \\ &=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \end{aligned}$
Hence it is proved.

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4

Answer: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$
Hints: First we will use the formula of $2 \tan ^{-1} x$.
Given: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$
Explanation:
$\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{3}<1 \end{array}\right]$
L.H.S:$\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) \end{aligned}$
$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{3}{28}<1 \end{array}\right]$
$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 5

Answer: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$
Hints: First we will convert $2 \tan ^{-1}\left(\frac{1}{3}\right)$in $\sin ^{-1}$
Given: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$
Explanation:
L.H.S:$\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right]$
$=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^{2}}\right)$ $\left[\begin{array}{c} -1 \leq x \leq 1 \\ -1 \leq \frac{1}{3} \leq 1 \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{\frac{10}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2}{3} \times \frac{9}{10}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{3}{5}\right) \end{aligned}$
$\left[\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right\}\right]$
$\text { [if } \left.x, y \geq 0 \text { and } x^{2}+y^{2} \leq 1\right]$
$\left[\begin{array}{l} \frac{4}{5}, \frac{3}{5} \geq 0,\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \\ \frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1 \leq 1 \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}\left\{\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right\} \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}+\frac{3}{5} \sqrt{1-\frac{16}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{\frac{16}{25}}+\frac{3}{5} \sqrt{\frac{9}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \times \frac{4}{5}+\frac{3}{5} \times \frac{3}{5}\right) \\ &=\sin ^{-1}\left(\frac{16}{25}+\frac{9}{25}\right) \\ &=\sin ^{-1}\left(\frac{25}{25}\right) \end{aligned}$
$\left[\begin{array}{l} \because \sin \frac{\pi}{2}=1 \\ \sin ^{-1}(1)=\frac{\pi}{2} \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}(1) \\ &=\frac{\pi}{2} \end{aligned}$
Hence it is proved that$\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 6

Answer: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$
Hint: First we will convert $\sin ^{-1}\left(\frac{3}{5}\right)$ into $\tan ^{-1}$.
Given: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$
Explanation:
Let us solve for $\sin ^{-1}\left(\frac{3}{5}\right)$
Let $\sin ^{-1}\left(\frac{3}{5}\right)=\theta$ ............(1)
$\begin{aligned} &\sin \theta=\left(\frac{3}{5}\right)=\frac{P}{H} \\ &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$
Now,
$\begin{aligned} &\tan \theta=\frac{P}{B} \\ &\tan \theta=\frac{3}{4} \\ &\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}$
From equation (1)
$\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$ .............(2)
L.H.S:
$2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$
from Equation (2)
$=2 \tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\begin{array}{c} -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \end{aligned}$
$\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} x \cdot y>-1 \\ \frac{24}{7} \times \frac{17}{31}=\frac{408}{217}>-1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{\frac{625}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$
$\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$
Hence it is Prove that $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7 .
Answer:

$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$
Hint: First we will solve for $2 \tan ^{-1} \frac{1}{5}$
Given: $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$
Explanation:
L.H.S:
$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$
$=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{5}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \end{aligned}$
$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{5}{12} \times \frac{1}{8}=\frac{5}{96}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{10+3}{24}}{1-\frac{5}{96}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right) \\ &=\tan ^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) \\ &=\tan ^{-1}\left(\frac{4}{7}\right) \end{aligned}$
Hence it is proved that $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 8 .

Answer:

$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$
Hint: First we will solve for $2 \tan ^{-1} \frac{3}{4}$
Given: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$
Explanation:
L.H.S:
$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1} \frac{17}{31} \end{aligned}$
$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y>-1 \\ \frac{24}{7} \times \frac{17}{31}>-1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that
$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 9.

Answer: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
Hint: First we will solve for _{$2 \tan ^{-1} \frac{1}{2}$}
Given: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
Explanation:
$2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{1}{1-\left(\frac{1}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{1}{3}{4}\right) \end{aligned}$
$2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{4}{3}\right)$ ...................(1)
L.H.S:
$2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$
From equation (1)
$=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{4}{3} \times \frac{1}{7}=\frac{4}{21}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) \\ &=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right) \\ &=\tan ^{-1}\left(\frac{31}{17}\right) \end{aligned}$
Hence it is proved that $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 10.

Answer: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$
Hints: First we will solve for $4 \tan ^{-1} \frac{1}{5}$
Split into $2 \times 2 \tan ^{-1} \frac{1}{5}$ and solve it
Given: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$
Explanation: Let us first solve _{$4 \tan ^{-1} \frac{1}{5}$}
_{$4 \tan ^{-1} \frac{1}{5}=2 \times 2 \tan ^{-1} \frac{1}{5}$}
_{$=2 \tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{5}<1 \end{array}\right]$}
_{$\begin{aligned} 4 \tan ^{-1} \frac{1}{5} &=2 \tan ^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right) \\ &=2 \tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right) \\ &=2 \tan ^{-1}\left(\frac{5}{12}\right) \end{aligned}$}
_{$=\tan ^{-1}\left(\frac{2 \times \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}\right) \quad\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<\frac{5}{12}<1 \end{array}\right]$}
_{$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{119}{144}}\right) \\ &=\tan ^{-1}\left(\frac{5}{6} \times \frac{144}{119}\right) \end{aligned}$}
_{$4 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{120}{119}\right)$} ......(1)
Now
L.H.S:
$\begin{aligned} &4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239} \\ &=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\ &=\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right) \end{aligned}$ $\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28680-119}{28441}}{\frac{28441+120}{28441}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{28561}{28441}}{\frac{28561}{28441}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$

Inverse Trigonomeric Functions Exercise 3.14 Question 3 .

Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}} \text { then } x=\frac{a-b}{1+a b}$
Hints: First we will convert whole L.H.S part in $\tan ^{-1}$
Given:$\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$
Explanation:
As we know that
$\begin{aligned} &2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ &2 \tan ^{1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$
Now $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$
$\begin{aligned} &2 \tan ^{-1} a-2 \tan ^{-1} b=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2\left(\tan ^{-1} a-\tan ^{-1} b\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x \\ &\frac{a-b}{1+a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a-b}{1+a b}=x \\ &x=\frac{a-b}{1+a b} \end{aligned}$
Hence it is proved that $x=\frac{a-b}{1+a b}$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1 .

Answer: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$
Hint: First we will convert $\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$ into $\tan ^{-1}$
Given: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$
Explanation:
L.H.S:
$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$
$=\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ $\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$
$=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right]$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right] \\ &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right] \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$
$=\tan ^{-1}(\infty)$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$
$=\frac{\pi}{2}$
Hence it is Proved that $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 2.

Answer: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$
Hint: First we will convert $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ into $\tan ^{-1}$
Given: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$
Explanation:
L.H.S:
$\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}$
$=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+2 \tan ^{-1} x\right\}$ $\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x\right]$
$=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\tan ^{-1} \frac{2 x}{1-x^{2}}\right\}$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$
$\begin{aligned} &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right\} \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$
$=\sin \left\{\tan ^{-1}(\infty)\right\}$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$
$\begin{aligned} &=\sin \left(\frac{\pi}{2}\right) \\ &=1 \end{aligned}$
Hence it is proved that $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 3.

Answer: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$
Hints: We will first convert $2 x \sqrt{1-x^{2}}$ into sin
Given: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$
$x=\sin \theta$
Let $\sqrt{1-x^{2}}=\sqrt{1-\sin ^{2} \theta}=\sqrt{\cos ^{2} \theta}=\cos \theta$ $\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right]$
Now
$\begin{aligned} &\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \end{aligned}$ $[2 \sin \theta \cos \theta=\sin 2 \theta]$
$\begin{aligned} &=2 \theta \\ &=2 \sin ^{-1} x \end{aligned}$ $\left[\begin{array}{l} \because \sin ^{-1}(\sin \theta)=\theta \\ x=\sin \theta \\ \theta=\sin ^{-1} x \end{array}\right]$
Hence it is proved that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$

Inverse Trigonomeric Functions Exercise 3.14 Question 5.

Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$to prove $x=\frac{a+b}{1-a b}$
Hints: First we will convert L.H.S of the question in $\tan ^{-1}$
Given: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$
Explanation:
$\begin{aligned} &\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \\ &\because 2 \tan ^{-1} a=\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right) \\ &\therefore \sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &2 \tan ^{-1} a+2 \tan ^{-1} b=2 \tan ^{-1} x \\ &2\left(\tan ^{-1} a+\tan ^{-1} b\right)=2 \tan ^{-1} x \\ &\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} x \end{aligned}$
$\tan ^{-1}\left(\frac{a+b}{1-a b}\right)=\tan ^{-1} x$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &\frac{a+b}{1-a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a+b}{1-a b}=x \end{aligned}$
Hence it is proved that $x=\frac{a+b}{1-a b}$

Inverse Trigonomeric Functions Exercise 3.14 Question 6.

Answer: $\pi$
Hint: First we will convert $\sin ^{-1} \frac{2 x}{1+x^{2}}$ into $\tan ^{-1}$
Given: $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$for $x \geq 1$
Explanation:
$2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right]$
$\begin{aligned} &=2 \tan ^{-1} x+2 \tan ^{-1} x \\ &=4 \tan ^{-1} x \quad(x \geq 1) \end{aligned}$
Let $x=1$
$\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
$\begin{aligned} &=4 \tan ^{-1}(1) \\ &=4 \times \frac{\pi}{4} \\ &=\pi \end{aligned}$
Hence the constant value for $x \geq 1$ is $\pi$

Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1.

Answer:

$\frac{\pi}{4}$
Given: Given that $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
Hint: First we solve $2 \sin ^{-1} \frac{1}{2}$ Since $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
Solution: We have,
$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
$=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}$ $\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right]$
$\begin{aligned} &=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\} \\ &=\tan ^{-1}\left\{2 \times \frac{1}{2}\right\} \end{aligned}$ $\left[\cos \frac{\pi}{3}=\frac{1}{2}\right]$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$
This is the required solution.

Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 2.

Answer: 0
Given: $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right),|x| \geq 1$
Hint: Since $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$applying it.
Solution: We have,
$\begin{aligned} &\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right) \\ &=\cos \left(\frac{\pi}{2}\right) \end{aligned}$ $\left[\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right]$
=0 $\left[\cos \frac{\pi}{2}=0\right]$
Hence, $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=0$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 1.

Answer: $x=\frac{-461}{9}$
Given: $\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
Hint: Use formula
$\text { 1. } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\text { 2. } 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Solution: We have,
$\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
$\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{4}+\frac{5}{12}}{1-\frac{1}{4} \times \frac{5}{12}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{8}{12}}{\frac{43}{48}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{32}{43}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \end{aligned}$
$\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1-\frac{32}{43} \times \frac{1}{6}}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ $\left[\tan ^{-1} \frac{\pi}{4}=1\right]$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{258}}{\frac{226}{258}}\right)+\tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\tan ^{-1}(1)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{235}{226}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}(1) \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\left(\frac{235}{226}\right) \times \frac{1}{x}}\right)=\tan ^{-1}(1), \frac{235}{226} x<1 \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{235 x+226}{226 x-235}=1, x>\frac{235}{226} \\ &\Rightarrow 235 x+226=226 x-235, x>\frac{235}{226} \\ &\Rightarrow 235 x-226 x=-235-226 \\ &\Rightarrow 9 x=-461 \\ &\Rightarrow x=\frac{-461}{9} \end{aligned}$
Hence $x=\frac{-461}{9}$is required answer

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2.

Answer:

$x=\frac{1}{\sqrt{3}}$
Given: $3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
Hint: Using formula since,
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution: We have,
$3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
We know that $2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\begin{aligned} &\Rightarrow 3\left(2 \tan ^{-1} x\right)-4\left(2 \tan ^{-1} x\right)+2\left(2 \tan ^{-1} x\right)=\frac{\pi}{3} \\ &\Rightarrow 6 \tan ^{-1} x-8 \tan ^{-1} x+4 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \end{aligned}$
$\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\right]$
$\Rightarrow x=\frac{1}{\sqrt{3}}$
This is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 3.

Answer: $x=\frac{1}{\sqrt{3}}$
Given: $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}, x>0$
Hint: Use
$\text { 1. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \text { and }$
$\text { 2. } \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution: We have,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}$
We know that,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ and convert $\cot ^{-1}$into $\tan ^{-1}$
$\Rightarrow 2 \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ $\left[\cot \theta=\frac{1}{\tan \theta}\right]$
$\Rightarrow 2 \tan ^{-1} x+2 \tan ^{-1} x=\frac{2 \pi}{3}$ $\left[\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow 4 \tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{2 \pi}{12} \\ &\Rightarrow x=\tan \frac{\pi}{6} \end{aligned}$ $\left[\tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}\right]$
$\Rightarrow x=\frac{1}{\sqrt{3}}$ $\left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 4.

Answer: $\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
Given: $2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x), x \neq \frac{\pi}{2}$
Hint: Using $2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Solution: We have,
$2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x)$
We know that
$2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}(2 \sec x) \\ &\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=2 \sec x \end{aligned}$ $\left[1-\sin ^{2} x=\cos ^{2} x\right]$
$\Rightarrow \frac{\sin x}{\cos x \cos x}=\frac{1}{\cos x}$ $\left[\sec x=\frac{1}{\cos x}\right]$
$\Rightarrow \tan x=1$ $\left[\frac{\sin x}{\cos x}=\tan x\right]$
$\begin{aligned} &\Rightarrow x=\tan ^{-1}(1) \\ &\Rightarrow x=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$
Hence the principal value of $x=n \pi+\frac{\pi}{4}, n \in Z$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 5.

Answer: $x=\sqrt{3}$
Given: $\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$
$\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution:
We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$
$\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
$\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \end{aligned}$
$\Rightarrow x=\sqrt{3}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$
Hence $x=\sqrt{3}$ is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 6.

Answer: $x=\sqrt{3}$
Given:$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$
$\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution:
We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$
$\begin{aligned} &\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1} x=\frac{2 \pi}{3} \end{aligned}$
$\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \\ &\Rightarrow x=\sqrt{3} \end{aligned}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$
Hence $x=\sqrt{3}$ is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 9.

Answer: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
Given: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
Hint: Using $2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution: Let us assume
$\begin{aligned} \text { L.H.S } &=2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right] \\ &=\cos ^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right] \end{aligned}$ $\left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$
$=\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}\right]$
$\begin{aligned} &=\cos ^{-1}\left[\frac{a+b-(a-b) \tan ^{2} \frac{\theta}{2}}{a+b+(a-b) \tan ^{2} \frac{\theta}{2}}\right] \\ &=\cos ^{-1}\left[\frac{a\left(1-\tan ^{2} \frac{\theta}{2}\right)+b\left(1+\tan ^{2} \frac{\theta}{2}\right)}{a\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right] \end{aligned}$
Dividing numerator and denominator by $\left(1+\tan ^{2} \frac{\theta}{2}\right)$ we get
$=\cos ^{-1}\left(\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)}\right]$
$=\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right]$ $\left[\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\cos \theta\right]$
$= R.H.S$
Hence $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$

Inverse Trigonometric Functions Exercise 3.14 Question 10.

Answer: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
Given: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$where $\alpha=a x-b y \: \: \& \: \: \beta=a y+b x$
Hint: Use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Solution: Let us assume
L.H.S:$=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\left(\frac{2 a b}{a^{2}-b^{2}}\right)\left(\frac{2 x y}{x^{2}-y^{2}}\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 a b x^{2}-2 a b y^{2}+2 x y a^{2}-2 x y b^{2}}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2\left(a b x^{2}+x y a^{2}-a b y^{2}-x y b^{2}\right)}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2[a x(b x+a y)-b y(a y+b x)]}{(a x-b y)^{2}-\left(a^{2} y^{2}+b^{2} x^{2}+2 a b x y\right)}\right] \end{aligned}$
$=\tan ^{-1}\left[\frac{2(b x+a y)(a x-b y)}{(a x-b y)^{2}-(b x+a y)^{2}}\right]$ $\left[(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
$=\tan ^{-1}\left[\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right]$ $[\alpha=a x-b y \& \beta=a y+b x]$
$=R.H.S$
Hence ,$\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$

Inverse Trigonomeric Functions Exercise 3.14 Question 11.

Answer: $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
Given: For any a, b, x, y > 0
We have to prove that $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
where,$\alpha=-a x+b y \& \beta=b x+a y$
Hint: First we will divide numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$respectively, then use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Solution: Let us assume
$\begin{aligned} L . H . S &=\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right) \\ &=\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 a b^{2}-a^{3}}{b^{3}}}{\frac{b^{3}-3 a^{2} b}{b^{3}}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 x y^{2}-x^{3}}{y^{3}}}{\frac{y^{3}-3 x^{2} y}{y^{3}}}\right] \end{aligned}$
Dividing numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$respectively.
$=\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{a}{b}\right)-\left(\frac{a}{b}\right)^{3}}{1-3\left(\frac{a}{b}\right)^{2}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{x}{y}\right)-\left(\frac{x}{y}\right)^{3}}{1-3\left(\frac{x}{y}\right)^{2}}\right]$
We know that,$3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
$\Rightarrow \mathrm{LH} . \mathrm{S}=\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right]+\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]$
$\begin{aligned} &=2 \tan ^{-1}\left(\frac{a}{b}\right)+2 \tan ^{-1}\left(\frac{x}{y}\right) \\ &=2\left[\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{x}{y}\right)\right] \end{aligned}$
$=2 \tan ^{-1}\left[\frac{\frac{a}{b}+\frac{x}{y}}{1-\frac{a}{b} \times \frac{x}{y}}\right]$ $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &=2 \tan ^{-1}\left[\frac{a y+b x}{b y-a x}\right] \\ &=2 \tan ^{-1}\left(\frac{\beta}{\alpha}\right) \end{aligned}$ $[b y-a x=\alpha \& \beta=a y+b x]$
$=\tan ^{-1}\left(\frac{2 \times \frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{2}}\right)$ $\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{2 \beta}{\alpha} \times \frac{\alpha^{2}}{\alpha^{2}-\beta^{2}}\right) \\ &=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) \\ &=\text { R.H.S } \end{aligned}$
Hence,$\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Where $\alpha=-a x+b y \& \beta=b x+a y$

RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 involves inverse trigonometric problems related to cosecant, secant, cosine, tangent functions. Solving these problems will make a student grasp better and perform well in exams.

In calculus, Inverse Trigonometric Functions are used to find various integrals. In addition to mathematics, inverse trigonometric functions are applied in science and engineering. Students will learn about the domain and range restrictions of trigonometric functions that ensure the existence of their inverses and how to observe their behavior using graphical representations and examples.

The specific exercise 3.14 has 31 questions, including subparts, and now a student can practice a lot and gain knowledge about the topic. In addition, this exercise will help in overall understanding of the chapter. RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 has practice questions which are to be solved and are important for exam point of view. The concepts are now easily understood by the students.

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